4
$\begingroup$

Is there a way to solve this integral? $$\int_{-\infty}^{\infty} e^{-(s^2+2\sqrt{t}s)}\sin(y+2\sqrt{t}s)\,ds$$ I usually apply this formula for integrals of this kind $$\int_{-\infty}^{\infty} e^{-(as^2+bs+c)}\,ds=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}-c}$$ Thank you in advance

$\endgroup$
  • 4
    $\begingroup$ Use $\sin(y+2\sqrt{t}s)=\frac{e^{i(y+2\sqrt{t}s)}-e^{-i(y+2\sqrt{t}s)}}{2i}$ and then use your formula. $\endgroup$ – aleden Oct 27 '18 at 17:42
  • 4
    $\begingroup$ With complex integration, this is easy (your last formula is valid for any complex $b$ - as both sides represent entire functions of $b$ - and that's enough to get the answer right away). $\endgroup$ – metamorphy Oct 27 '18 at 17:42
6
$\begingroup$

If you're OK with working with functions of a complex variable, the formula you gave still applies if $a,b,c\in\mathbb C$ provided that $\mathrm{Re}[a] > 0$. It gives \begin{multline} \int_{-\infty}^{\infty} e^{-(s^2+2\sqrt{t}s)}\sin(y+2\sqrt{t}s)\,ds = \mathrm{Im}\left[\int_{-\infty}^{\infty} e^{-(s^2+2\sqrt{t}s)}e^{i\left(y+2\sqrt{t}s\right)}ds\right] \\= \mathrm{Im}\left[\int_{-\infty}^{\infty} e^{-[s^2+2(1-i)\sqrt{t}s-iy]}ds\right] = \mathrm{Im}\left[\sqrt{\pi}\exp\left(-2it +i y\right)\right] = \sqrt{\pi}\sin(y-2t). \end{multline} If you're not OK with working with complex numbers, there actually are formulas for these: \begin{eqnarray} \int_{-\infty}^\infty e^{-(as^2 + bs + c)}\sin(ks + m)ds &=& \sqrt{\frac{\pi}{a}}e^{\frac{b^2-k^2}{4a} - c}\sin\left(m-\frac{b k}{2a}\right)\\ \int_{-\infty}^\infty e^{-(as^2 + bs + c)}\cos(ks + m)ds &=& \sqrt{\frac{\pi}{a}}e^{\frac{b^2-k^2}{4a} - c}\cos\left(m-\frac{b k}{2a}\right) \end{eqnarray} These formulas follow directly from the complex form of the integral I used before, and if you plug in the values, you'll get $\sqrt{\pi}\sin(y-2t)$ again.

If you want to know how to get these formulas without using complex numbers, well, that's a more involved post.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.